Linear Algebra – Exercise 13, Section 2.C of Linear Algebra Done Right

linear algebrasolution-verificationvector-spaces

Exercise: Suppose $U$ and $W$ are both $4$-dimensional subspaces of $C^6$. Prove that
there exist two vectors in $U \cap W$ such that neither of these vectors is a
scalar multiple of the other.

My attempt at a proof is as follows.

Proof: Let $u_1,. . .,u_4$ be a basis of $U$ and let $w_1,. . .,w_4$ be a basis of $W$. Then, $u_1,. . .,u_4,w_1,. . .,w_4$ spans $U+W$. Because $U+W$ is a subspace of $C^6$, the $\dim(U+W)\le 6$. Thus, $u_1,. . .,u_4,w_1,. . .,w_4$ can be reduced to a basis of $U+W$. In the process, none of the $u's$ get removed as $u_1,. . .,u_4$ is linearly independent. Thus, some of the $w's$ get removed in the process. Because $\dim(U+W)\le 6$, at least two of the $w's$ get removed. These are the $w's\in U\cap W$. Because $w_1, . .,w_4$ is linearly independent, none of these two vectors are a scalar multiple of each other.

Is the proof correct?

Edit: I implicitly use that theorem that every spanning list in a vector space can be reduced to a basis of that vector space. In the process, we remove those vectors that are in the span of the previous ones. Thus, if we have the list $v_1,. . .,v_k$. We remove $v_j$ only if $v_j$ is in the span of $v_1,. . .,v_{j-1}$.

Edit 2: I have come to know that the proof is wrong. For future readers, I am writing another proof that is also suggested as a hint in the answers.

Proof 2: Using the formula $\dim (U+W)=\dim(U)+\dim(W)-\dim(U\cap W)$, we see that $\dim(U\cap W) \ge 2$. This is because $\dim(C^6)=6$ and $U+W$ is a subspace of $C^6$. Thus, $\dim(U+W)\le 6$. Let $j\in Z^+$ with $2\le j\le 6$. Let $\dim(U\cap W)=j$. Let $v_1,. . .,v_j$ be a basis of $U\cap W$. Then we have that $v_1,v_2\in U\cap W$ are not scalar multiples of each other as they are linearly independent. Completing the proof.

Best Answer

If by $w'$ you mean some of the $w_1, \dots, w_4$, then no, your proof is not correct. Those of $w_1, \dots, w_4$ which you remove do not have to lie in $U$. It's easy to construct an example where none of $w_1, \dots, w_4$ lie in $U$. For example, let $e_1, \dots, e_6$ be a basis of $\mathbb{C}^6$; $u_i=e_i$ for $i=1,\dots, 4$; $w_1 = e_5$; $w_2=e_6$; $w_3=e_1+e_5$; $w_4=e_2+e_6$.

Related Question